EXAMPLE 3.2 - Using the grammar in Example 32, show a parse tree and a leftmost derivation for the following statement: A=Aast (B+(Cast A)) A Grammar for Simple Assignment Statements cassign>-dd>-cexpn <ib=Al B |C capp-cid>+expp | ( <expp) I dd

To show a parse tree and a leftmost derivation for the statement , we can use the grammar provided in Example 32. The grammar for simple assignment statements is as follows:``` ::= = ::= | ::= ::= +```Here, `` represents an assignment statement, `` represents an identifier, `` represents an expression, `` represents an application of two expressions, and `` represents an operator.To construct a parse tree, we start with the given statement and apply the grammar rules to break it down into its components. The parse tree for the statement would look like this:```A├── A = A * (B + (C * A))│ ├── A│ ├── *│ ├── B + (C * A)│ │ ├── B│ │ ├── +│ │ ├── C * A│ │ │ ├── C│ │ │ ├── *│ │ │ ├── A```To construct a leftmost derivation, we start with the given statement and apply the grammar rules to generate a sequence of symbols that represents the statement. The leftmost derivation for the statement would look like this:```A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A => A (B + (C * A)) => A * + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (BC * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A)) => A = A * (B + (C * A))